Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $a = \dfrac{5k^3 - 80k^2 + 300k}{-2k^3 - 2k^2 + 40k} \div \dfrac{k - 6}{3k + 15} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{5k^3 - 80k^2 + 300k}{-2k^3 - 2k^2 + 40k} \times \dfrac{3k + 15}{k - 6} $ First factor out any common factors. $a = \dfrac{5k(k^2 - 16k + 60)}{-2k(k^2 + k - 20)} \times \dfrac{3(k + 5)}{k - 6} $ Then factor the quadratic expressions. $a = \dfrac {5k(k - 6)(k - 10)} {-2k(k + 5)(k - 4)} \times \dfrac {3(k + 5)} {k - 6} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac { 5k(k - 6)(k - 10) \times 3(k + 5)} { -2k(k + 5)(k - 4) \times (k - 6)} $ $a = \dfrac {15k(k - 6)(k - 10)(k + 5)} {-2k(k + 5)(k - 4)(k - 6)} $ Notice that $(k + 5)$ and $(k - 6)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {15k(k - 6)(k - 10)\cancel{(k + 5)}} {-2k\cancel{(k + 5)}(k - 4)(k - 6)} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $a = \dfrac {15k\cancel{(k - 6)}(k - 10)\cancel{(k + 5)}} {-2k\cancel{(k + 5)}(k - 4)\cancel{(k - 6)}} $ We are dividing by $k - 6$ , so $k - 6 \neq 0$ Therefore, $k \neq 6$ $a = \dfrac {15k(k - 10)} {-2k(k - 4)} $ $ a = \dfrac{-15(k - 10)}{2(k - 4)}; k \neq -5; k \neq 6 $